A integral inequality and generalization If $ f(x)$ is decreasing then $\int_{a-1}^{b}f(x)dx\ge \sum_{k=a}^b f(k)\ge \int_a^{b+1}f(x)dx$.
And reverse for increasing functions.
I try to generalize it
Generalization: 
Let $f(x)$ be a monotone decreasing function on $\left(0,+\infty\right)$. Prove that:
$$ \int_{a}^{b+1}f(x)dx\leq  \sum_{k=a}^{b}f(k)\leq \int_{a-1}^{b}f(x)dx\leq \sum_{k=a-1}^{b-1}f(k)\quad a,b\in\mathbb N$$
When does equality hold?
*Note that $0\notin \mathbb N$
I have 3 questions as follow : 
1) Could you please prove generalization 
2) Is this true ? Generalization (inversion)
Let $f(x)$ be a monotone increasing function on $\left(0,+\infty\right)$. Prove that:
$$ \int_{a}^{b+1}f(x)dx\geq  \sum_{k=a}^{b}f(k)\geq \int_{a-1}^{b}f(x)dx\geq \sum_{k=a-1}^{b-1}f(k)\quad a,b\in\mathbb N$$
When does equality hold?
3) Can we have a stronger generalization ?
Thank you in advance! 
 A: 1) Once you have the fact $$\int_{a-1}^{b}f(x)dx\ge \sum_{k=a}^b f(k)\ge \int_a^{b+1}f(x)dx,\ \ \forall a,b\in \mathbb N,$$ there is not much to prove, for at the expression $$ \int_{a}^{b+1}f(x)dx\leq  \sum_{k=a}^{b}f(k)\leq \int_{a-1}^{b}f(x)dx\leq \sum_{k=a-1}^{b-1}f(k),\ \forall a,b\in\mathbb N,$$ the first two inequalities follow directly from the fact and the third one follows from the fact applied to $a-1$ and $b-1$ in the place of $a$ and $b$, respectively.
2) The proof for increasing functions is analog to that for decreasing functions. Using the fact for increasing functions
$$\int_{a-1}^{b}f(x)dx\le \sum_{k=a}^b f(k)\le \int_a^{b+1}f(x)dx,\ \ \forall a,b\in \mathbb N,$$
at the expression $$ \int_{a}^{b+1}f(x)dx\geq  \sum_{k=a}^{b}f(k)\geq \int_{a-1}^{b}f(x)dx\geq \sum_{k=a-1}^{b-1}f(k),\ \forall a,b\in\mathbb N,$$ the first two inequalities follow directly from the fact and the third one follows from the fact applied to $a-1$ and $b-1$ in the place of $a$ and $b$, respectively.
3) I don't know and I can't think of any suggestions.
About the question of when the equality holds for (1), I have an idea. Note that
$\displaystyle\int_{a-1}^{b}f(x)dx = \int_a^{b+1}f(x)dx$ if, and only if,
$$
\int_{a-1}^{a}f(x)dx + \int_{a}^{b}f(x)dx = \int_a^{b}f(x)dx + \int_{b}^{b+1}f(x)dx\ \ \Leftrightarrow\ \ \int_{a-1}^{a}f(x)dx = \int_{b}^{b+1}f(x)dx.
$$
So our problem is equivalent to analyze which conditions imply that
$$
\int_{a-1}^{a}f(x)dx = \int_{b}^{b+1}f(x)dx.
$$
The simplest condition that guarantees the equality above is:

Condition 1. $a-1=b$;

Other two conditions for which the equality immediately holds are:

Condition 2. $a-1<b$ and $f$ is a constant at the interval $(a-1,b+1)$;
Condition 3. $a-1>b$ and $f$ is a constant at the interval $(b,a)$;

I guess these are the onliest cases for that the equality holds, the proof is not so difficult.
A: For the second inequality note that $\int_{a-1}^b f(x) dx = \sum_{k=a-1}^{b-1} \int_k^{k+1} f(x) dx$. Now by monotonicity we get that $\int_k^{k+1} f(x) dx \geq f(k+1)$ and hence the second inequality.
For the third inequality we can use the same reasoning as above except that we use $\int_k^{k+1} f(x) dx \leq f(k)$.
As stated above the equality in the first inequality holds if we have a piecewise constant function which is constant on $[k, k+1)$ for $k = a, \dots, b$. Now assume that we have a function $g$ that satisfies the equality. Then $g$ is piecewise constant on $[k, k+1)$ for $k = a, \dots, b$. If it was not then there would exist a $k \in \{a, \dots, b\}$ and some $\xi \in [k, k+1)$ for which we have $g(k) > g(\xi)$ since $g$ is decreasing. This would however imply that $\int_k^{k+1} g(x) dx < g(k)$ and hence we do not have equality.
